Translating Graphs | 图像的平移

📚 Translating Graphs | 图像的平移

Translating a graph is one of the simplest yet most important transformations in A-Level mathematics. By adding constants to the input or output of a function, we can shift the entire curve horizontally, vertically, or both without altering its size, shape, or orientation. This topic underpins many areas of calculus, coordinate geometry, and trigonometric analysis, and is regularly assessed on Edexcel papers.

图像的平移是A-Level数学中最简单但也最重要的变换之一。通过向函数的输入或输出添加常数，我们可以将整条曲线水平移动、垂直移动或同时移动，而不改变其大小、形状或方向。这一主题是微积分、坐标几何和三角分析的基础，并在Edexcel考试中频繁出现。

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1. Introduction to Graph Translations | 图像平移简介

A translation moves every point of a graph by the same amount in a given direction. In function notation, if we start with y = f(x), then the translated graph can be written as y = f(x – h) + k, where h and k are real numbers. The constants h and k control the horizontal and vertical shifts respectively.

平移会在给定方向上将图上的每一个点移动相同的距离。在函数符号中，如果我们从 y = f(x) 开始，那么平移后的图像可以写成 y = f(x – h) + k，其中 h 和 k 是实数。常数 h 和 k 分别控制水平位移和垂直位移。

It is essential to recognise that inside the function, x – h shifts the graph to the right when h > 0, and to the left when h < 0. Outside the function, +k shifts the graph upward for k > 0, and downward for k < 0. This reverse behaviour of the horizontal shift often confuses students, so careful attention is needed.

必须认识到，在函数内部，x – h 在 h > 0 时使图像向右平移，h < 0 时向左平移。而在函数外部，+k 在 k > 0 时使图像向上平移，k < 0 时向下平移。水平位移的这种反向特性常常使学生困惑，因此需要格外注意。

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2. Vertical Translations: f(x) + a | 垂直平移：f(x) + a

The simplest translation is a vertical shift. Replacing f(x) by f(x) + a adds a to every y-coordinate of the original graph. If a is positive, the graph moves up; if a is negative, it moves down. The shape of the graph, its x-intercepts (unless the graph is moved across the x-axis), and its intervals of increase and decrease are preserved, but the y-intercept changes.

最简单的平移是垂直位移。将 f(x) 替换为 f(x) + a 时，原图每一个点的 y 坐标都加上 a。如果 a 为正，图像向上移动；如果 a 为负，图像向下移动。图形的形状、单调增减区间保持不变，但 y 截距会改变，而 x 截距可能改变（除非图像跨过 x 轴）。

Vertical translation: y = f(x) + a

For example, the graph of y = x² shifted up by 3 units becomes y = x² + 3. The vertex (0,0) moves to (0,3), and the axis of symmetry remains x = 0.

例如，y = x² 向上平移 3 个单位得到 y = x² + 3。顶点 (0,0) 移动到 (0,3)，对称轴仍为 x = 0。

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3. Horizontal Translations: f(x + a) | 水平平移：f(x + a)

A horizontal translation is achieved by replacing x with x ± a inside the function. The transformation y = f(x + a) shifts the graph to the left by a units if a > 0, and to the right by |a| units if a < 0. This is opposite to what many students initially expect, because the shift is applied to the input variable x.

水平平移是通过在函数内部将 x 替换为 x ± a 来实现的。变换 y = f(x + a) 在 a > 0 时将图像向左平移 a 个单位，在 a < 0 时向右平移 |a| 个单位。这与许多学生最初的直觉相反，因为位移是作用于输入变量 x 的。

Horizontal translation: y = f(x + a)

For the quadratic y = (x – 2)², the graph of y = x² is shifted 2 units to the right. All key features such as the minimum point and axis of symmetry move accordingly.

对于二次函数 y = (x – 2)²，y = x² 的图像被向右平移了 2 个单位。所有关键特征，如极小值点和对称轴，都随之移动。

To understand why, consider a specific point. The original graph of y = f(x) passes through (c, f(c)). After the translation x → x – 2, the new function is y = f(x – 2). Then when x = c + 2, we get y = f(c), so the point (c, f(c)) has moved to (c + 2, f(c)). Hence the whole curve shifts right by 2.

要理解原因，可以考虑一个具体的点。原图像 y = f(x) 经过点 (c, f(c))。经过平移 x → x – 2 后，新函数为 y = f(x – 2)。当 x = c + 2 时，我们得到 y = f(c)，因此点 (c, f(c)) 移动到了 (c + 2, f(c))。所以整个曲线向右移动了 2 个单位。

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4. Combining Horizontal and Vertical Translations | 水平和垂直平移的结合

When both horizontal and vertical shifts are applied, the function takes the general form y = f(x – h) + k. The graph of f(x) is translated h units horizontally and k units vertically. The order in which the translations are applied does not matter because addition is commutative – shifting right then up produces the same result as shifting up then right.

当同时进行水平和垂直移动时，函数取一般形式 y = f(x – h) + k。f(x) 的图像被水平平移 h 个单位并垂直平移 k 个单位。由于加法满足交换律，平移的顺序无关紧要——先向右再向上与先向上再向右得到的结果相同。

y = f(x – h) + k

For instance, the graph of y = √x can be transformed to y = √(x – 3) + 1. The original graph is shifted right by 3 and up by 1. The starting point of the radical (originally at (0,0)) moves to (3,1).

例如，y = √x 的图像可以变换为 y = √(x – 3) + 1。原图向右平移 3 个单位，向上平移 1 个单位。根式函数的起点（原在 (0,0)）移动到 (3,1)。

A common mistake is to write y = f(x – h) + k but then interpret the horizontal shift as moving left for positive h. Always test with a known point: if h = 2, then the value f(0) now occurs at x = 2, confirming a right shift.

一个常见的错误是写出了 y = f(x – h) + k，却将水平位移理解为 h 为正时向左移动。始终应用已知点进行检验：如果 h = 2，那么 f(0) 的值现在出现在 x = 2，这证实了是向右平移。

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5. Effect on Key Points and Asymptotes | 对关键点和渐近线的影响

Under the translation y = f(x – h) + k, every point (x₀, y₀) on the original graph is mapped to (x₀ + h, y₀ + k). This rule applies to stationary points, intercepts, and any other feature. For rational functions, vertical asymptotes x = a move to x = a + h, and horizontal asymptotes y = b move to y = b + k.

在平移 y = f(x – h) + k 下，原图像上的每一个点 (x₀, y₀) 都被映射到 (x₀ + h, y₀ + k)。这条规则适用于驻点、截距以及任何其他特征。对于有理函数，竖直渐近线 x = a 移动到 x = a + h，水平渐近线 y = b 移动到 y = b + k。

For example, the hyperbola y = 1/x has asymptotes at x = 0 and y = 0. Translating by vector (2, –1) gives y = 1/(x – 2) – 1, with asymptotes x = 2 and y = –1.

例如，双曲线 y = 1/x 的渐近线为 x = 0 和 y = 0。按向量 (2, –1) 平移后得到 y = 1/(x – 2) – 1，渐近线变为 x = 2 和 y = –1。

This principle helps when sketching graphs without plotting numerous points: locate the transformed key features first, then draw the original shape in its new position.

这一原则有助于在不绘制大量点的情况下草图：先确定平移后的关键特征位置，然后在新位置绘制原始形状。

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6. Translation of Quadratic Functions | 二次函数的平移

Quadratics are a natural setting to practise translations. Starting with the basic parabola y = x², the translated form is y = (x – h)² + k. This is exactly the completed-square form of a quadratic, where the vertex is (h, k). Thus, a translation can turn y = x² into any congruent parabola with a vertical axis of symmetry.

二次函数是练习平移的自然背景。从基本的抛物线 y = x² 出发，平移后的形式为 y = (x – h)² + k。这正是二次函数的完全平方形式，其顶点为 (h, k)。因此，平移可以将 y = x² 变成任何具有竖直对称轴的合同抛物线。

For example, translating y = x² left 4 and down 5 gives y = (x + 4)² – 5. Here h = –4 and k = –5, so the vertex is (–4, –5). The line of symmetry becomes x = –4.

例如，将 y = x² 向左平移 4 个单位再向下平移 5 个单位得到 y = (x + 4)² – 5。这里 h = –4，k = –5，因此顶点为 (–4, –5)。对称轴变为 x = –4。

If the quadratic is given in standard form such as y = x² – 6x + 4, we can complete the square to y = (x – 3)² – 5, revealing that the graph is a translation of y = x² shifting right 3 and down 5.

如果二次函数以标准形式给出，例如 y = x² – 6x + 4，我们可以通过配方法得到 y = (x – 3)² – 5，从而揭示该图像是 y = x² 向右平移 3 个单位再向下平移 5 个单位的结果。

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7. Translation of Trigonometric Functions | 三角函数的平移

Trigonometric graphs undergo translations in exactly the same way. The function y = sin(x – α) represents a horizontal translation (phase shift) of y = sin x by α units to the right. Adding a constant D, as in y = sin(x) + D, shifts the midline vertically. These transformations are critical when modelling periodic phenomena.

三角函数的图像以完全相同的方式发生平移。函数 y = sin(x – α) 表示将 y = sin x 向右平移 α 个单位（相移）。添加一个常数 D，如 y = sin(x) + D，会使中线垂直移动。这些变换在模拟周期现象时至关重要。

For a complete translation, we may write y = sin(x – α) + D. The graph of y = sin x has amplitude 1, period 360° and oscillates around y = 0. After translation, the amplitude and period are unchanged, but the central line becomes y = D, and the graph is shifted horizontally so that a maximum originally at 90° moves to 90° + α.

对于完整的平移，我们可以写成 y = sin(x – α) + D。y = sin x 的图像振幅为 1，周期为 360°，并围绕 y = 0 振荡。平移后，振幅和周期保持不变，但中线变为 y = D，并且图像水平移动，原本在 90° 的最大值移动到 90° + α。

For instance, y = sin(x – 30°) + 1 shifts the sine wave right by 30° and lifts it up by 1. Key points like (0°, 0) become (30°, 1), (90°, 1) become (120°, 2), etc.

例如，y = sin(x – 30°) + 1 将正弦波向右移动 30° 并向上提升 1 个单位。关键点如 (0°, 0) 变成 (30°, 1)，(90°, 1) 变成 (120°, 2)，依此类推。

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8. Translating Exponential and Logarithmic Functions | 指数函数和对数函数的平移

Exponential graphs of the form y = aˣ (a > 0) have a horizontal asymptote at y = 0. Under a vertical translation y = aˣ + k, the asymptote shifts to y = k. A horizontal translation y = aˣ⁻ʰ simply moves the entire curve left or right, but the asymptote remains unchanged because exponential functions approach y = 0 as x → –∞.

指数函数 y = aˣ (a > 0) 的图像有一条水平渐近线 y = 0。在垂直平移 y = aˣ + k 下，渐近线移至 y = k。水平平移 y = aˣ⁻ʰ 只是将整个曲线向左或向右移动，但渐近线保持不变，因为指数函数在 x → –∞ 时接近 y = 0。

Logarithmic functions such as y = ln x have a vertical asymptote at x = 0. The translation y = ln(x – h) + k moves the asymptote to x = h, while the horizontal shift changes the domain to x > h. For example, y = ln(x – 2) + 1 has an asymptote at x = 2 and a domain of x > 2.

对数函数如 y = ln x 有一条竖直渐近线 x = 0。平移 y = ln(x – h) + k 将渐近线移到 x = h，同时水平位移将定义域变为 x > h。例如，y = ln(x – 2) + 1 的渐近线为 x = 2，定义域为 x > 2。

When sketching these, always draw the shifted asymptote first. Then plot a few transferred points to guide the shape of the curve.

在画草图时，始终先画出平移后的渐近线。然后标出几个转换后的点来指导曲线的形状。

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9. Finding the Image of a Point Under a Translation | 求点在平移下的像

Given a specific point (p, q) that lies on y = f(x), the corresponding point on y = f(x – h) + k is (p + h, q + k). This is a powerful tool for constructing translated graphs without a calculator. The mapping works for any function, regardless of complexity.

给定位于 y = f(x) 上的一个特定点 (p, q)，那么在 y = f(x – h) + k 上相应的点为 (p + h, q + k)。这是一个在没有计算器的情况下构建平移图像的强大工具。该映射适用于任何函数，无论其复杂程度如何。

Example: The point (2, –3) lies on y = g(x). Find the image point on y = g(x + 5) – 4. Here, rewrite as g(x – (–5)) – 4, so h = –5, k = –4. The image is (2 – 5, –3 – 4) = (–3, –7).

示例：点 (2, –3) 位于 y = g(x) 上。求在 y = g(x + 5) – 4 上的像点。这里可重写为 g(x – (–5)) – 4，因此 h = –5，k = –4。像点为 (2 – 5, –3 – 4) = (–3, –7)。

This method avoids the common error of applying the wrong sign. Always express the translation in the form (x – h) to read h directly, even if it is negative.

该方法避免了符号应用错误的常见问题。始终将平移表示为 (x – h) 的形式，以便直接读出 h，即使 h 为负值亦然。

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10. Using Vector Notation for Translations | 使用向量符号表示平移

In Edexcel A-Level Mathematics, translations are often expressed using column vectors. A translation that moves a graph a units horizontally and b units vertically is written as translation by vector (a b), where the top entry is the horizontal shift and the bottom entry is the vertical shift. The notation is widely used in coordinate geometry and transformation questions.

在Edexcel A-Level数学中，平移通常用列向量表示。将一个图像水平移动 a 个单位、垂直移动 b 个单位的平移记为按向量 (a b) 平移，其中上方分量是水平位移，下方分量是垂直位移。该符号在坐标几何和变换题中广泛使用。

For a function y = f(x), the effect of translation by vector (h k) is to produce the new function y = f(x – h) + k. Hence, the signs inside the brackets follow the same rule: a positive h in the vector means shift right, and mathematically appears as (x – h).

对于函数 y = f(x)，按向量 (h k) 平移的效果是产生新函数 y = f(x – h) + k。因此，括号内的符号遵循同样的规则：向量中的 h 为正表示向右平移，并在数学上以 (x – h) 的形式出现。

Students are expected to recognise and use both the functional form and the vector form interchangeably. A typical exam question might state: “Describe the transformation that maps y = f(x) onto y = f(x – 3) + 2.” The answer: translation by vector (3 2).

学生应能识别并交替使用函数形式和向量形式。典型的考题可能会问：“描述将 y = f(x) 映射到 y = f(x – 3) + 2 的变换。”答案是：按向量 (3 2) 平移。

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11. Solving Equations Involving Translated Graphs | 解涉及平移图像的方程

When a function is translated, the solutions to f(x) = c are translated in the same way. If the original equation f(x) = 0 has roots x = α, β, then the transformed function f(x – h) + k = 0 can be solved by first solving f(X) = –k, where X = x – h, and then adjusting by h.

当函数发生平移时，方程 f(x) = c 的解也以相同的方式平移。如果原方程 f(x) = 0 有根 x = α, β，则变换后的函数 f(x – h) + k = 0 可通过先求解 f(X) = –k（其中 X = x – h）然后加上 h 来求解。

For example, given that f(x) = x² – 4x + 3 has roots at x = 1 and x = 3, find the roots of f(x – 2) + 1 = 0. Set X = x – 2, then f(X) = –1. The original f(x) = –1 leads to x = ? We can simply shift the roots: the new equation solves to (x – 2) = original solutions of f(t) = –1. But a quicker insight: the entire graph is shifted right by 2 and up by 1, so the x-intercepts of the translated graph correspond to the shifted intercepts of f(x) = –1.

例如，已知 f(x) = x² – 4x + 3 有根 x = 1 和 x = 3，求 f(x – 2) + 1 = 0 的根。设 X = x – 2，则 f(X) = –1。我们可以直接通过平移求解：新方程的根是原方程 f(t) = –1 的解加上 2。对于这个具体的 f，我们首先解 x² – 4x + 3 = –1，得到 x = 2。因此平移后唯一根为 x = 2 + 2 = 4（因为重复根）。

While this particular case led to a double root, the approach is general: solve the “inner” equation for the shifted variable, then reverse the shift.

虽然这个特殊情况产生了重根，但方法具有普遍性：先解出平移变量满足的“内部”方程，然后再反向回代位移量。

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12. Common Mistakes and Tips | 常见错误与提示

Mistake 1: Forgetting the direction of horizontal shifts. Many students write y = f(x – 2) and think it moves left. Always check: substituting x = 2 gives f(0), so the original y-intercept has moved to x = 2 – a right shift. Use a simple point test on every problem.

错误 1：忘记水平位移的方向。许多学生写下 y = f(x – 2) 却以为图像向左移动。始终检验：代入 x = 2 得到 f(0)，所以原 y 轴截距已移到 x = 2 处——这是向右移动。在每个问题中使用简单的点测试。

Mistake 2: Applying translations to the wrong variable. A translation written as f(x + 3) + 2 must not be interpreted as f(x) + 3 + 2. The +3 is inside the function argument, not added to the result.

错误 2：将平移应用到了错误的变量上。写成 f(x + 3) + 2 的平移不能被解释

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